Question

A frame {B}, initially coinciding with frame {A}, is first rotated of 45° about the xaxis of frame {A} and afterward rotated of 90° about the y-axis of frame {A}. Finally, the so obtained frame {B}’’ is translated of 10 units about its own y-axis. Find the homogeneous transform matrix of the final rototranslated frame {B}’’’ relative to frame {A}.

A frame {B}, initially coinciding with frame {A}, is first rotated of 45° about the x- axis of frame {A} and afterward rotated of 90° about the y-axis of frame {A}. Finally, the so obtained frame {B}" is translated of 10 units about its own y-axis. Find the homogeneous transform matrix of the final rototranslated frame {B}" relative to frame {A}.

Answer 1

First Consider rotations:

$\small ^AT_{B'' }= R_{A,y}(90^0)*R_{A,x}(45^0)$

We can write this using the standard x and y rotation matrices.

$\small \Rightarrow ^AT_{B''} = \begin{bmatrix} cos90^0 & 0 &sin90^0&0 \\ 0 & 1 &0&0 \\ -sin90^0 & 0 & cos90^0&0\\0&0&0&1 \end{bmatrix}*\begin{bmatrix} 1 & 0 &0&0 \\ 0 & cos45^0 & -sin45^0&0\\ 0 & sin45^0 & cos45^0 &0 \\ 0&0&0&1\end{bmatrix}$

= 4 TB" = 0 0 1 0 0 1 0 0 -1 0 0 0 0 0 0 1 1.0000 0 0 0 0.7071 0.7071 0 0 0 -0.7071 0 0.7071 0 0 1

+4 TB" = B" 0 0 -1 0 0.7071 0.7071 0 0 0.70710 -0.7071 0 0 0 0 1

Now We consider translation with respect to the B'' frame by 10 units about its own y-axis.

For this, we post-multiply a translation transformation matrix to the above matrix.

$\small \Rightarrow ^AT_{B'''} = \begin{bmatrix} 0&0.7071 & 0.7071 & 0\\ 0&0.7071 & -0.7071 & 0 \\-1 &0&0 &0\\ 0&0&0&1 \end{bmatrix}* \begin{bmatrix} 1 & 0 & 0& 0\\ 0 & 1 & 0 & 10\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix}$

$\small \Rightarrow ^AT_{B'''} = \begin{bmatrix} 0 & 0.7071 & 0.7071 & 7.0711\\ 0 & 0.7071& -0.7071 & 7.0711\\ -1& 0 & 0 & 0\\ 0 & 0 & 0 &1 \end{bmatrix}$